\(\int F^{a+b (c+d x)^n} (c+d x)^{-1+n} \, dx\) [373]

Optimal result

Integrand size = 23, antiderivative size = 27 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+n} \, dx=\frac {F^{a+b (c+d x)^n}}{b d n \log (F)} \]

[Out]

F^(a+b*(d*x+c)^n)/b/d/n/ln(F)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {2240} \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+n} \, dx=\frac {F^{a+b (c+d x)^n}}{b d n \log (F)} \]

[In]

Int[F^(a + b*(c + d*x)^n)*(c + d*x)^(-1 + n),x]

[Out]

F^(a + b*(c + d*x)^n)/(b*d*n*Log[F])

Rule 2240

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^n*(
F^(a + b*(c + d*x)^n)/(b*f*n*(c + d*x)^n*Log[F])), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {F^{a+b (c+d x)^n}}{b d n \log (F)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+n} \, dx=\frac {F^{a+b (c+d x)^n}}{b d n \log (F)} \]

[In]

Integrate[F^(a + b*(c + d*x)^n)*(c + d*x)^(-1 + n),x]

[Out]

F^(a + b*(c + d*x)^n)/(b*d*n*Log[F])

Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04

method	result	size
risch	\(\frac {F^{a +b \left (d x +c \right )^{n}}}{b d n \ln \left (F \right )}\)	\(28\)
norman	\(\frac {{\mathrm e}^{\left (a +b \,{\mathrm e}^{n \ln \left (d x +c \right )}\right ) \ln \left (F \right )}}{d b n \ln \left (F \right )}\)	\(32\)

[In]

int(F^(a+b*(d*x+c)^n)*(d*x+c)^(n-1),x,method=_RETURNVERBOSE)

[Out]

F^(a+b*(d*x+c)^n)/b/d/n/ln(F)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+n} \, dx=\frac {e^{\left ({\left (d x + c\right )}^{n} b \log \left (F\right ) + a \log \left (F\right )\right )}}{b d n \log \left (F\right )} \]

[In]

integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1+n),x, algorithm="fricas")

[Out]

e^((d*x + c)^n*b*log(F) + a*log(F))/(b*d*n*log(F))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 104 vs. \(2 (19) = 38\).

Time = 3.08 (sec) , antiderivative size = 104, normalized size of antiderivative = 3.85 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+n} \, dx=\begin {cases} \frac {x}{c} & \text {for}\: F = 1 \wedge b = 0 \wedge d = 0 \wedge n = 0 \\F^{a} \left (\frac {c \left (c + d x\right )^{n - 1}}{d n} + \frac {x \left (c + d x\right )^{n - 1}}{n}\right ) & \text {for}\: b = 0 \\F^{a + b c^{n}} c^{n - 1} x & \text {for}\: d = 0 \\\frac {F^{a + b} \log {\left (\frac {c}{d} + x \right )}}{d} & \text {for}\: n = 0 \\\frac {c \left (c + d x\right )^{n - 1}}{d n} + \frac {x \left (c + d x\right )^{n - 1}}{n} & \text {for}\: F = 1 \\\frac {F^{a + b \left (c + d x\right )^{n}}}{b d n \log {\left (F \right )}} & \text {otherwise} \end {cases} \]

[In]

integrate(F**(a+b*(d*x+c)**n)*(d*x+c)**(-1+n),x)

[Out]

Piecewise((x/c, Eq(F, 1) & Eq(b, 0) & Eq(d, 0) & Eq(n, 0)), (F**a*(c*(c + d*x)**(n - 1)/(d*n) + x*(c + d*x)**(
n - 1)/n), Eq(b, 0)), (F**(a + b*c**n)*c**(n - 1)*x, Eq(d, 0)), (F**(a + b)*log(c/d + x)/d, Eq(n, 0)), (c*(c +
d*x)**(n - 1)/(d*n) + x*(c + d*x)**(n - 1)/n, Eq(F, 1)), (F**(a + b*(c + d*x)**n)/(b*d*n*log(F)), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+n} \, dx=\frac {F^{{\left (d x + c\right )}^{n} b + a}}{b d n \log \left (F\right )} \]

[In]

integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1+n),x, algorithm="maxima")

[Out]

F^((d*x + c)^n*b + a)/(b*d*n*log(F))

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+n} \, dx=\frac {F^{{\left (d x + c\right )}^{n} b + a}}{b d n \log \left (F\right )} \]

[In]

integrate(F^(a+b*(d*x+c)^n)*(d*x+c)^(-1+n),x, algorithm="giac")

[Out]

F^((d*x + c)^n*b + a)/(b*d*n*log(F))

Mupad [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int F^{a+b (c+d x)^n} (c+d x)^{-1+n} \, dx=\frac {F^{a+b\,{\left (c+d\,x\right )}^n}}{b\,d\,n\,\ln \left (F\right )} \]

[In]

int(F^(a + b*(c + d*x)^n)*(c + d*x)^(n - 1),x)

[Out]

F^(a + b*(c + d*x)^n)/(b*d*n*log(F))